Atkins' Physical Chemistry
Atkins' Physical Chemistry
11th Edition
ISBN: 9780198769866
Author: ATKINS, P. W. (peter William), De Paula, Julio, Keeler, JAMES
Publisher: Oxford University Press
Question
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Chapter 18, Problem 18C.5AE

(i)

Interpretation Introduction

Interpretation:

Entropy of activation has to be calculated for second order gas-phase ozone decomposition which has frequency factor as 4.6x1012dm3mol-1s-1 and activation energy of 10.0kJmol-1.

(i)

Expert Solution
Check Mark

Explanation of Solution

Given information:

    A= 4.6x1012dm3mol-1s-1T= 298KEa= 10.0kJmol-1

Entropy of activation:

In the problem statement it is said that the reaction takes place at low pressure.  Hence, the reaction is assumed to be bimolecular and the rate constant is second-order.

The rate constant equation can also be given as shown below by use of Eyring equation.

    kr = BeΔS/ReΔH/RTWhere, B = (kTh)(RTpO) = kRT2hpO

Comparing the above equation with equation (1),

    kr = BeΔS/ReEa/RTe= Ae(Ea/RT)

Therefore.

  A= eBeΔS/R

The above equation implies that,

    ΔS = R(lnAB2)(1)

B can be calculated as shown below,

    B = kRT2hpO = (1.381x10-23JK-1)x(8.314JK-1mol-1)x(298K)26.626x10-34Jsx105Pa = 1019615.025x10-23J2mol-16.626x10-34Jsx1atm = 1.538x1016Jatm-1mol-1s-1

It is known that 1J is equal to 9.87x10-3dm3atm.  Therefore, the value of B can be expressed as shown below,

    B = 1.538x1016x9.87x10-3dm3atmatm-1mol-1s-1 = 15.180x1013dm3mol-1s-1 = 1.518x1014dm3mol-1s-1

Substitution of the obtained values in the equation (1), entropy of activation can be calculated as shown below,

    ΔS = R(lnAB2) = R(ln4.6x1012dm3mol-1s-11.518x1014dm3mol-1s-12) = 8.314JK-1mol-1(5.496) = 45.69JK-1mol-1

The entropy of activation is calculated as 45.69JK-1mol-1.

(ii)

Interpretation Introduction

Interpretation:

Enthalpy of activation has to be calculated for second order gas-phase ozone decomposition which has frequency factor as 4.6x1012dm3mol-1s-1 and activation energy of 10.0kJmol-1.

(ii)

Expert Solution
Check Mark

Explanation of Solution

Given information:

    A= 4.6x1012dm3mol-1s-1T= 298KEa= 10.0kJmol-1

Enthalpy of activation:

Enthalpy of activation can be given by the equation shown below,

  Ea = ΔH+RTΔH = EaRT = 10.0kJmol-1(8.314x10-3kJK-1mol-1x298K) = 10.0kJmol-1(8.314x10-3kJmol-1x298) = 10.0kJmol-12.477kJmol-1 = 7.523kJmol-1

Therefore, the enthalpy of activation is calculated as 7.523kJmol-1.

(iii)

Interpretation Introduction

Interpretation:

Gibbs energy of activation has to be calculated for second order gas-phase ozone decomposition which has frequency factor as 4.6x1012dm3mol-1s-1 and activation energy of 10.0kJmol-1.

(iii)

Expert Solution
Check Mark

Explanation of Solution

Given information:

    A= 4.6x1012dm3mol-1s-1T= 298KEa= 10.0kJmol-1

Enthalpy of activation (ΔH) is calculated as 7.523kJmol-1.

Entropy of activation (ΔS) is calculated as 45.69JK-1mol-1.

Gibbs energy of activation:

Gibbs energy can be calculated using the equation given below,

    ΔG = ΔHTΔS(1)

Substitution of the obtained values in equation (1), Gibbs energy of activation can be calculated as shown below,

    ΔG = 7.523kJmol-1(298Kx(45.69JK-1mol-1)) = 7.523kJmol-1+13.61562kJK-1mol-1 = 21.138kJmol-1

Therefore, the Gibbs energy of activation is calculated as 21.138kJmol-1.

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Chapter 18 Solutions

Atkins' Physical Chemistry

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